Bulgarian Mathematical Competitions

Solution:
It is easy to see that if the points $A$, $I$, $J$ and $C$ are collinear, then $AB = AD$ and $BC = CD$. Hence $ABCD$ is a circumscribed quadrilateral.

Suppose that the points $A$, $I$, $J$ and $C$ are concyclic. Since $\angle AIC > \angle AIB$ or $\angle AIC > \angle AID$, it follows that $\angle AIC > 90^\circ$. Analogously $\angle AJC > 90^\circ$ and therefore $\angle AIC + \angle AJC > 180^\circ$. It follows that the points $I$ and $J$ are on the same side of the line $AC$.

Let $S = AI \cap CJ$ and let the lines $AI$ and $CJ$ meet the circumcircle of $ABCD$ at points $P$ and $Q$, respectively. Since $P$ and $Q$ are the midpoints of the arcs $BCD$ and $BAD$, respectively, it follows that $PQ \perp BD$.

On the other hand, $\angle SJI = \angle CAI = \angle CQP$, which implies that $IJ \parallel PQ$. Therefore $IJ \perp BD$, which means that the incircles of $\triangle ABD$ and $\triangle BCD$ are tangent to each other at a point $T \in BD$. Then
$$DT = \frac{AD + BD - AB}{2} = \frac{BD + CD - BC}{2} \Longleftrightarrow AB + CD = BC + AD$$
i.e. $ABCD$ is a circumscribed quadrilateral.

Conversely, assume that $ABCD$ is a circumscribed quadrilateral. Note that if $I \in AC$, then $J \in AC$. Suppose that $I \notin AC$. It follows from the equality $AB + CD = BC + AD$ that the incircles of $\triangle ABD$ and $\triangle BCD$ are tangent to each other at a point of $BD$. Hence $IJ \perp BD \Rightarrow IJ \parallel PQ \Rightarrow \angle SJI = \angle CQP = \angle CAI$, and therefore $ACJI$ is a cyclic quadrilateral.